Answer
$\sin(\theta)\tan(\theta)+\cos(\theta)=\sec(\theta)$
Work Step by Step
$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$
Therefore,
$\sin(\theta)\tan(\theta)+\cos(\theta)=\sin(\theta)\frac{\sin(\theta)}{\cos(\theta)}+\cos(\theta)=\frac{\sin^{2}(\theta)}{\cos(\theta)}+\cos(\theta)$
To obtain a common denominator, $\cos(\theta)$ is multiplied by $\frac{\cos(\theta)}{\cos(\theta)}$,
$\frac{\sin^{2}(\theta)}{\cos(\theta)}+\cos(\theta)=\frac{\sin^{2}(\theta)+\cos^{2}(\theta)}{\cos(\theta)}$
Using the identity $\sin^{2}(\theta)+\cos^{2}(\theta)=1$,
$\frac{\sin^{2}(\theta)+\cos^{2}(\theta)}{\cos(\theta)}=\frac{1}{\cos(\theta)}=\sec(\theta)$
Therefore,
$\sin(\theta)\tan(\theta)+\cos(\theta)=\sec(\theta)$
